Question

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the centre of the solenoid and 3.50 cm from its axis is 8.00 x 10-6 V/m. Calculate di/dt

Short answer

If a long thin solenoid having 400 turns per meter and radius of , the current in the solenoid is increasing at a unform rate of 9.21A/s

Step-by-step solution

Calculating the value of E

It is given that a long solenoid of 400 turns and distance of ${\mathrm{m}}^{-1}$and a radius of 1.10cm . The induced electric field at a point near the centre of the solenoid is 3.50cm and induced electric field is $8.00\times {10}^{-6}\mathrm{V}/\mathrm{m}$

Now using Faraday’s law we get:

$$\begin{gathered} \epsilon =\oint \overrightarrow{E}.\overrightarrow{dl} \\ =-\frac{d\varphi B}{dt} \end{gathered}$$

Now, the magnetic flux through the circle of radius r > R :

$$\begin{gathered} \mathrm{\varphi }=\mathrm{AB} \\ ={\mathrm{\pi R}}^2\mathrm{B} \end{gathered}$$

Now integrating LHS we get:

$$\begin{gathered} \oint \overrightarrow{E}.\overrightarrow{dl} \\ =E\left(2\pi r\right) \end{gathered}$$

Absolute value of E is:

$E\left(2\pi r\right)=\pi R^2\left|\frac{dB}{dt}\right|$

We can also write it as:

$E=\frac{R^2{\mu }_{0}n}{2r}=\left|\frac{dB}{dt}\right|$

Calculating the values

Now to get $\frac{dl}{dt}$we get:

$\left|\frac{\mathrm{dl}}{\mathrm{dt}}\right|=\frac{2\mathrm{Er}}{{\mathrm{\mu }}_0{\mathrm{nR}}^2}$

Now, substituting the values we get:

$$\begin{gathered} \frac{2\left(8.00\times {10}^{-6}\mathrm{V}/\mathrm{m}\right)\left(0.0350\mathrm{m}\right)}{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\left(400\right){\left(0.0110\mathrm{m}\right)}^2} \\ =9.21\mathrm{A}/\mathrm{s} \end{gathered}$$

Therefore, the value of$\left|\frac{dl}{dt}\right|$ is = 9.21A/s